What is the integral of 9^{x+1} ?

The integral of 9^{x+1} is (9^{x+1})/(2ln(3))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 9^{x+1}

\int 9^{x + 1} d x

\frac{9 ^{x + 1}}{2 ln ( 3 )} + C

Solution steps

\int 9^{x + 1} d x

Simplify

9^{x + 1} : 9^{x} \cdot 9

= \int 9^{x} \cdot 9 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int 9^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 9 \cdot \frac{9 ^{x}}{ln ( 9 )}

Simplify

9 \cdot \frac{9 ^{x}}{ln ( 9 )} : \frac{9 ^{x + 1}}{2 ln ( 3 )}

= \frac{9 ^{x + 1}}{2 ln ( 3 )}

Add a constant to the solution

= \frac{9 ^{x + 1}}{2 ln ( 3 )} + C

\int e^{x} \frac{1 + x ^{2}}{( x + 1 ) ^{2}} d x

\int \frac{x e ^{x}}{2 ( 1 + x ) ^{2}} d x

\int (4 x + 4) d x

\int \frac{1}{1 + x} + \frac{1}{1 - x} d x

\int 2^{9 - x} - 64 d x

What is the integral of 9^{x+1} ?
The integral of 9^{x+1} is (9^{x+1})/(2ln(3))+C